房间错误 - 如何解决？

Question

我在我的 android 应用程序上使用了持久性空间，一旦我创建了我的数据库类，我就会收到以下错误：

• 错误：实体类必须用@Entity注解
• 错误：实体和 Pojo 必须有一个可用的公共构造函数。您可以有一个空的构造函数或一个参数与字段匹配的构造函数（按名称和类型）。
• 错误：实体必须至少有 1 个用 @PrimaryKey 注释的字段

我将其缩小到我的数据库类，我消除了我的实体和DAO 类以外的所有其他内容。我已经尝试了我在网上找到的所有内容，但没有任何效果。

这是实体类：

import android.arch.persistence.room.ColumnInfo;
import android.arch.persistence.room.Entity;
import android.arch.persistence.room.ForeignKey;
import android.arch.persistence.room.PrimaryKey;
import android.content.pm.PackageManager;

import static android.arch.persistence.room.ForeignKey.CASCADE;

@Entity
public class Character {

@PrimaryKey(autoGenerate = true)
public int characterId;

@ColumnInfo(name = "name")
public String name;

@ColumnInfo(name = "race")
public String race;

@ColumnInfo(name = "occupation")
public String occupation;

@ColumnInfo(name = "shadow")
public String shadow;

@ColumnInfo(name = "accurate")
public int accurate;

@ColumnInfo(name = "cunning")
public int cunning;

@ColumnInfo(name = "discreet")
public int discreet;

@ColumnInfo(name = "persuasive")
public int persuasive;

@ColumnInfo(name = "quick")
public int quick;

@ColumnInfo(name = "resolute")
public int resolute;

@ColumnInfo(name = "strong")
public int strong;

@ColumnInfo(name = "vigilant")
public int vigilant;

@ColumnInfo(name = "toughness")
public int toughness;

@ColumnInfo(name = "painThreshold")
public int painThreshold;

@ColumnInfo(name = "corruptionThreshold")
public int corruptionThreshold;

@ColumnInfo(name = "defense")
public int defense;

public int inventoryId;

public int spellBookId;

public Character() {

}

public Character(int characterId, String name, String race, String occupation, String shadow, int accurate, int cunning, int discreet, int persuasive, int quick, int resolute, int strong, int vigilant, int toughness, int painThreshold, int corruptionThreshold, int defense, int inventoryId, int spellBookId) {
    this.characterId = characterId;
    this.name = name;
    this.race = race;
    this.occupation = occupation;
    this.shadow = shadow;
    this.accurate = accurate;
    this.cunning = cunning;
    this.discreet = discreet;
    this.persuasive = persuasive;
    this.quick = quick;
    this.resolute = resolute;
    this.strong = strong;
    this.vigilant = vigilant;
    this.toughness = toughness;
    this.painThreshold = painThreshold;
    this.corruptionThreshold = corruptionThreshold;
    this.defense = defense;
    this.inventoryId = inventoryId;
    this.spellBookId = spellBookId;
}
...getters and setters

这是数据库类：

@Database (entities = {Character.class}, version = 1)
public abstract class AppDatabase extends RoomDatabase {

public abstract CharDao charDao();
}

我的 DAO 类：

@Dao
public interface CharDao {

@Insert
void addChar (Character character);

}

Build.gradle：

apply plugin: 'com.android.application'

android {
compileSdkVersion 28
defaultConfig {
    applicationId "symbhero.studio.com.symbhero"
    minSdkVersion 15
    targetSdkVersion 28
    versionCode 1
    versionName "1.0"
    testInstrumentationRunner 
"android.support.test.runner.AndroidJUnitRunner"
}
buildTypes {
    release {
        minifyEnabled false
        proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
        }
    }
}

dependencies {
implementation fileTree(dir: 'libs', include: ['*.jar'])
implementation 'com.android.support:appcompat-v7:28.0.0'
implementation 'com.android.support.constraint:constraint-layout:1.1.3'
testImplementation 'junit:junit:4.12'
androidTestImplementation 'com.android.support.test:runner:1.0.2'
androidTestImplementation 'com.android.support.test.espresso:espresso-core:3.0.2'
implementation 'android.arch.persistence.room:runtime:1.1.0'
annotationProcessor 'android.arch.persistence.room:compiler:1.1.0'
}

我真的不知道我哪里出错了。我按照link 进行了所有设置。

感谢所有帮助！

Answer 1

您必须检查所有与 ROOM 相关的类 [@Entity、@Dao、@Database] 是否必须
保存在同一个包中。有时您需要手动从应用程序文件夹中删除构建
从资源管理器中重新构建代码。

Answer 2

一切看起来都很好。但我想你可能会在 DAO 或 AppDatabase 类中导入不同的类而不是 @Entity 字符。

很少有编码技巧。

• 如果您在 @ColumnInfo 中定义与字段名称相同的名称，则不需要 @ColumnInfo 注释。
• 如果您有 public 变量，那么您不必有 setter 和 getter

Answer 3

也许我的意见可以帮助你:)

实体：

1. 列表项

缺少表名：

@Entity(tableName = "your_tabel_name_here")

1. 列表项

没有列信息的变量

public int inventoryId;

public int spellBookId;

1. 列表项

忽略除空构造函数之外的所有构造函数

@Ignore

public Character(int characterId, String name, String race ...)

**DAO：**

1. 首先，将 Dao 命名为与 Entity 相同：CharacterDAO

2. 使用以下插入样式：

   @道 公共接口 CharacterDAO {

   @Insert(onConflict = OnConflictStrategy.IGNORE)
   void insertCharacter(Character... character);
   

   }

**应用数据库：**

@Database(entities = {Character.class},version = 1, exportSchema = false)
public abstract class AppDatabase extends RoomDatabase {

    private static AppDatabase INSTANCE;

    public abstract CharacterDAO characterDAO();
private static String DATABASE_NAME = "your_database_name";

    //https://developer.android.com/reference/android/arch/persistence/room/RoomDatabase.html
    public static synchronized AppDatabase getAppDatabase(Context context) {
        if (INSTANCE == null) {
            INSTANCE =
                    Room.databaseBuilder(context.getApplicationContext(), AppDatabase.class, DATABASE_NAME)
                            .build();
        }
        return INSTANCE;
    }